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设a1=2,a2=4,数列{bn}满足:bn=an+1-an,bn+1=2bn+2.(1)求b1、b2;(2)求证数列{bn+2}是等比数列(要指出首项与公比);(3)求数列{an}的通项公式.
题目详情
设a1=2,a2=4,数列{bn}满足:bn=an+1-an,bn+1=2bn+2.
(1)求b1、b2;
(2)求证数列{bn+2}是等比数列(要指出首项与公比);
(3)求数列{an}的通项公式.
(1)求b1、b2;
(2)求证数列{bn+2}是等比数列(要指出首项与公比);
(3)求数列{an}的通项公式.
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答案和解析
(1)b1=a2-a1=4-2=2,b2=2b1+2=2×2+2=6.
(2)证明:∵bn+1=2bn+2,∴bn+2+2=2(bn+2).
∴数列{bn+2}是以b1+2=4为首项,2为公比的等比数列.
(3)由(2)可得:bn+2=4×2n−1=2n+1.
∴bn=2n+1−2.
∴an−an−1=2n−2.
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=(2n-2)+(2n-1-2)+…+(22-2)+2
=2n+2n-1+…+22+2-2(n-1)
=
-2n+2
=2n+1-2n.
(2)证明:∵bn+1=2bn+2,∴bn+2+2=2(bn+2).
∴数列{bn+2}是以b1+2=4为首项,2为公比的等比数列.
(3)由(2)可得:bn+2=4×2n−1=2n+1.
∴bn=2n+1−2.
∴an−an−1=2n−2.
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=(2n-2)+(2n-1-2)+…+(22-2)+2
=2n+2n-1+…+22+2-2(n-1)
=
2(2n−1) |
2−1 |
=2n+1-2n.
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